ΑD=AF=x ,FC=CE=y,BE=BD=z .if a+b+c=2t We know that x=t-a,y=t-c,z=t-bS1=π(t-a)^2/8 so √[8S1/π]=t-a similarly √[8S2/π]=t-c so (8/π)√(S1S2)=(t-a)t-c)S=√t(t-a)(t-c)(t-b) but easily seen (t-a)t-c)=√t(t-a)(t-c)(t-b)because a^2+c^2=b^2solution by Michael Tsourakakis from Greece